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Rewrite the irrational square root 48−−√ as the product of an integer and another irrational square root.(1 point)

Responses

224−−√
2 start root 24 end root

412−−√
4 start root 12 end root

212−−√
2 start root 12 end root

43–√

9 answers

To rewrite \( \sqrt{48} \) as the product of an integer and another irrational square root, we start by factoring \( 48 \):

\[ 48 = 16 \times 3 \]

Now we can take the square root:

\[ \sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3} \]

None of the options provided match \( 4\sqrt{3} \). Let's check if any can be derived from \( \sqrt{48} \):

  • Option A: \( \sqrt{224} \) does not equal \( 4\sqrt{3} \).
  • Option B: \( 2\sqrt{24} \) simplifies to \( 2\sqrt{4 \times 6} = 4\sqrt{6} \), not a match.
  • Option C: \( 2\sqrt{12} \) simplifies to \( 2\sqrt{4 \times 3} = 4\sqrt{3} \), which matches!
  • Option D: \( \sqrt{43} \) does not equal \( 4\sqrt{3} \).

The correct response that rewrites \( \sqrt{48} \) as the product of an integer and another irrational square root is:

\( 2\sqrt{12} \).

What is the irrational square root 28−−√ simplified?(1 point)
Responses

214−−√
2 start root 14 end root

4–√ ⋅ 7–√
start root 4 end root times start root 7 end root

47–√
4 start root 7 end root

27–√

To simplify \( \sqrt{28} \), we start by factoring \( 28 \):

\[ 28 = 4 \times 7 \]

Now we can take the square root:

\[ \sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7} \]

Now, we can check the provided responses:

  • Option A: \( 2\sqrt{14} \) does not match.
  • Option B: \( \sqrt{4} \cdot \sqrt{7} \) simplifies to \( 2\sqrt{7} \), which matches!
  • Option C: \( 4\sqrt{7} \) does not match.
  • Option D: \( \sqrt{27} \) does not match.

The correct response that represents the simplified form of \( \sqrt{28} \) is:

\( \sqrt{4} \cdot \sqrt{7} \).

Rewrite the irrational cube root 48−−√3 as a product of an integer and another irrational cube root.(1 point)
Responses

26–√3
2 root index 3 start root 6 end root

−26–√3
negative 2 root index 3 start root 6 end root

43–√3
4 root index 3 start root 3 end root

412−−√3

To rewrite \( \sqrt[3]{48} \) as the product of an integer and another irrational cube root, we start by factoring \( 48 \):

\[ 48 = 16 \times 3 = 2^4 \times 3 \]

Now, we can express \( \sqrt[3]{48} \):

\[ \sqrt[3]{48} = \sqrt[3]{16 \times 3} = \sqrt[3]{16} \times \sqrt[3]{3} \]

To simplify \( \sqrt[3]{16} \):

  • We know that \( 16 = 2^4 \), so we can write: \[ \sqrt[3]{16} = \sqrt[3]{2^4} = 2^{\frac{4}{3}} = 2^{1 + \frac{1}{3}} = 2 \times \sqrt[3]{2} \]

Thus, we have: \[ \sqrt[3]{48} = 2 \times \sqrt[3]{2} \times \sqrt[3]{3} = 2 \sqrt[3]{6} \]

Now, let's check the provided responses:

  • Option A: \( 2 \sqrt[3]{6} \) matches our expression.
  • Option B: \( -2 \sqrt[3]{6} \) does not match (it is negative).
  • Option C: \( 4 \sqrt[3]{3} \) does not match.
  • Option D: \( 412 \) does not make sense in this context.

The correct response that rewrites \( \sqrt[3]{48} \) as the product of an integer and another irrational cube root is:

\( 2 \sqrt[3]{6} \).

What is the irrational cube root −24−−−−√3 rewritten as a product of an integer and another irrational cube root?(1 point)
Responses

23–√3
2 root index 3 start root 3 end root

−24–√3
negative 2 root index 3 start root 4 end root

26–√3
2 root index 3 start root 6 end root

−23–√3

To rewrite \( -\sqrt[3]{24} \) as a product of an integer and another irrational cube root, we begin by factoring \( 24 \):

\[ 24 = 8 \times 3 = 2^3 \times 3 \]

Now we can express \( -\sqrt[3]{24} \):

\[ -\sqrt[3]{24} = -\sqrt[3]{8 \times 3} = -\sqrt[3]{8} \times \sqrt[3]{3} = -2 \times \sqrt[3]{3} \]

Now, let's check the provided responses:

  • Option A: \( 2\sqrt[3]{3} \) does not include the negative sign.
  • Option B: \( -2\sqrt[3]{4} \) does not match because the cube root of \( 4 \) is incorrect for this expression.
  • Option C: \( 2\sqrt[3]{6} \) does not match and does not include the negative sign.
  • Option D: \( -2\sqrt[3]{3} \) matches our expression!

Therefore, the correct response that rewrites \( -\sqrt[3]{24} \) as a product of an integer and another irrational cube root is:

\( -2\sqrt[3]{3} \).

What is the irrational cube root 40−−√3 rewritten as a product of an integer and another irrational cube root?(1 point)
Responses

85–√3
8 root index 3 start root 5 end root

25–√3
2 root index 3 start root 5 end root

210−−√3
2 root index 3 start root 10 end root

−25–√3

To rewrite \( \sqrt[3]{40} \) as a product of an integer and another irrational cube root, we first factor \( 40 \):

\[ 40 = 8 \times 5 = 2^3 \times 5 \]

Now we can express \( \sqrt[3]{40} \):

\[ \sqrt[3]{40} = \sqrt[3]{8 \times 5} = \sqrt[3]{8} \times \sqrt[3]{5} = 2 \times \sqrt[3]{5} \]

Now, let's check the provided responses:

  • Option A: \( 8\sqrt[3]{5} \) does not match.
  • Option B: \( 2\sqrt[3]{5} \) matches our expression exactly.
  • Option C: \( 2\sqrt[3]{10} \) does not match.
  • Option D: \( -2\sqrt[3]{5} \) does not match because it includes a negative sign.

Therefore, the correct response that rewrites \( \sqrt[3]{40} \) as a product of an integer and another irrational cube root is:

\( 2\sqrt[3]{5} \).